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posted by  sylvia~ on 11/2/2009 10:30:01 PM  |  status: Closed  |  Earned Karma: 33

gcd( ) function

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N/A N/A N/A N/A 11/4/2009 at 11:00:00 PM
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a) Let a be a positive integer. Show that gcd(a,a-1) = 1.
b) Use the result of part a) to solve the Diophantine equation a+b=ab,
where (a,b) are positive integers.
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posted by A-ManESL on 11/3/2009 3:03:12 AM  |  status: Live
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(a) The gcd of a,b is the least positive value of Xa + Yb. Clearly as 1.a + (-1).(a-1) = 1 so (a,a-1) = 1.
 
(b) We need to solve a + b = ab. This is true iff a = b(a-1) which is true iff (a-1) divides a. In view of (a-1,a) = 1 this is true iff a-1 = 1 or a = 2. But then substituting in a + b = ab tells us that b = 2 as well. 
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posted by sylvia~ on 11/3/2009 3:50:48 AM  |  status: Live
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(a) The gcd of a,b is the least positive value of Xa + Yb. Clearly as 1.a + (-1).(a-1) = 1 so (a,a-1) = 1.
gcd is "greatest common divisor" ranther than the least positive value. I dont understand your solution. Can you explain? Thx.
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posted by A-ManESL on 11/3/2009 7:39:05 AM  |  status: Live
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One of the most important results concerning gcd is the following. This is so important that sometimes it is taken as the definition of the gcd.
Result: If g is the gcd of a and b then there exists integers X and Y such that g = Xa + Yb. Also g is the least positive number expressible in this form xa + yb.
Proof: Consider the set S = {xa+yb: x,y are integers}. Choose d to be the least positive integer of this set, so that d = Xa + Yb for some integers X and Y. We shall now show that d = g. For this we shall first show that d divides both a and b. We'll in fact show only that d divides a. The other case is similar.
To prove that d divides a assume the contrary is true, i.e. d does not divide a. Then application of the division algorithm for d and a gives a non zero remainder, i.e. a = dq + r where 0 < r < d. Hence we have r = a - dq = a - q(Xa + Yb) = (1-qX)a + (-qY)b. So r is a positive number in S by virtue of having the desired form of elements in S. But then 0 < r < d contradicts the fact that d was chosen to be the least positive integer of S. So r must be 0 and so d must divide a.
Next we show that g = d. Note that d is a common divisor of a and b and so d ≤ g by definition of g. But as g divides both a and b it also divides their linear combinations so that g divides Xa + Yb i.e. g divides d. Both integers being positive this implies that g ≤ d. Hence g = d.
By virtue of this result you are entitled to think of the gcd as the least positive number among all numbers having the form xa + yb. Clearly if any combination of a and b is 1 that is the least positive possible and so obviously their gcd is 1. This is what happens in the case of a and a-1.
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